Home › Free Resources › Maths Worksheets › IB Maths AA SL – Topic 1: Number and Algebra
Free Maths Worksheet

IB Maths AA SL – Topic 1: Number and Algebra

Sequences and series, exponents and logarithms, binomial theorem, proof.

WhatsApp Mr. AhmadBook a trial lesson (50% off)Free worksheets
12 questions  ·  64 marks  ·  answers included  ·  IB AA SL
Download the PDFAsk about a question

This worksheet covers:

  • Arithmetic and geometric sequences and series
  • Exponents and logarithms
  • The binomial theorem and simple deductive proof
  • In the style of IB AA SL Paper 1
  • Paper 2

Section A is non-calculator; allow about 75 minutes for the full set.

Questions

Section A – Skills

Non-calculator
1

An arithmetic sequence has first term $7$ and common difference $4$.(a) Find the 20th term.(b) Find the sum of the first 20 terms.

Show answer
(a) $u_{20} = 7 + 19(4) = 83$(b) $S_{20} = \frac{20}{2}(7 + 83) = 900$
[4]
2

A geometric sequence has $u_1 = 3$ and $u_2 = 6$.(a) Write down the common ratio.(b) Find $u_8$.(c) Find the sum of the first 8 terms.

Show answer
(a) $r = 2$(b) $u_8 = 3 \times 2^7 = 384$(c) $S_8 = \frac{3(2^8 – 1)}{2 – 1} = 765$
[5]
3

Solve for $x$.(a) $2^{3x-1} = 16$(b) $\log_x 49 = 2$(c) $\log_2 40 – \log_2 5 = x$

Show answer
(a) $3x – 1 = 4 \Rightarrow x = \frac{5}{3}$(b) $x^2 = 49 \Rightarrow x = 7$ (base must be positive)(c) $\log_2 8 = 3$, so $x = 3$
[6]
4

(a) Find the coefficient of $x^3$ in the expansion of $(2 + x)^5$.(b) Find the constant term in the expansion of $\left(x – \dfrac{2}{x}\right)^6$.

Show answer
(a) $\binom{5}{3}2^2 = 10 \times 4 = 40$(b) Term $\binom{6}{3}x^3\left(-\frac{2}{x}\right)^3 = 20(-8) = -160$
[5]
5

The geometric series $18 + 12 + 8 + \dots$ converges.(a) Find the common ratio and state why the series converges.(b) Find the sum to infinity.

Show answer
(a) $r = \frac{2}{3}$; converges because $|r| < 1$(b) $S_\infty = \dfrac{18}{1 – \frac{2}{3}} = 54$
[4]
6

Evaluate $\displaystyle\sum_{k=1}^{20}(3k + 2)$.

Show answer
Arithmetic series with $u_1 = 5$, $u_{20} = 62$: $S_{20} = \frac{20}{2}(5 + 62) = 670$
[3]

Section B – Exam-style

Calculator allowed
7

Maya invests $5000$ USD in an account paying $4.2\%$ per annum, compounded quarterly.(a) Find the value of the investment after 6 years.(b) Find how many years it takes for the investment to double in value.

Show answer
(a) $5000\left(1 + \frac{0.042}{4}\right)^{24} = 6424.53$ USD (accept $6420$ USD)(b) $\left(1.0105\right)^{4n} = 2 \Rightarrow n = \dfrac{\ln 2}{4\ln 1.0105} = 16.6$ years (16.58; so during the 17th year)
[6]
8

In an arithmetic sequence, $u_4 = 17$ and $u_{10} = 41$.(a) Find the common difference and the first term.(b) Find the least value of $n$ for which $S_n > 1000$.

Show answer
(a) $6d = 24 \Rightarrow d = 4$, $u_1 = 5$(b) $S_n = \frac{n}{2}(10 + 4(n-1)) = n(2n + 3)$; $2n^2 + 3n – 1000 > 0 \Rightarrow n > 21.6$, so $n = 22$ ($S_{22} = 1034$, $S_{21} = 945$)
[6]
9

A geometric sequence has $u_2 = 6$ and $u_5 = 48$.(a) Find the common ratio and the first term.(b) Find $S_{10}$.(c) Find the smallest $n$ such that $u_n > 10\,000$.

Show answer
(a) $r^3 = 8 \Rightarrow r = 2$, $u_1 = 3$(b) $S_{10} = 3(2^{10} – 1) = 3069$(c) $3 \times 2^{n-1} > 10000 \Rightarrow n – 1 > \log_2(3333.3) = 11.7$, so $n = 13$
[7]
10

Solve $\log_2 x + \log_2(x – 2) = 3$.

Show answer
$\log_2\big(x(x-2)\big) = 3 \Rightarrow x^2 – 2x – 8 = 0 \Rightarrow (x-4)(x+2) = 0$; $x = 4$ only ($x = -2$ rejected since $x > 2$ required)
[5]

Section C – Challenge

11

In the expansion of $(1 + 2x)^n$, where $n \in \mathbb{Z}^+$, the coefficient of $x^2$ is $112$.(a) Show that $n = 8$.(b) Hence find the coefficient of $x^3$.

Show answer
(a) $\binom{n}{2}2^2 = 112 \Rightarrow \frac{n(n-1)}{2} = 28 \Rightarrow n^2 – n – 56 = 0 \Rightarrow (n-8)(n+7) = 0$, so $n = 8$(b) $\binom{8}{3}2^3 = 56 \times 8 = 448$
[6]
12

(a) Prove that $(n + 3)^2 – (n – 3)^2 = 12n$ for all $n \in \mathbb{Z}$.(b) Prove that the sum of the squares of any two consecutive odd integers is 2 more than a multiple of 8.

Show answer
(a) LHS $= n^2 + 6n + 9 – (n^2 – 6n + 9) = 12n = $ RHS(b) Let the integers be $2k+1$ and $2k+3$: $(2k+1)^2 + (2k+3)^2 = 8k^2 + 16k + 10 = 8(k^2 + 2k + 1) + 2$, which is 2 more than a multiple of 8
[7]

Related worksheets

More worksheets in IB Mathematics: Analysis & Approaches:

IB Maths AA SL – Topic 2: FunctionsIB Maths AA SL – Topic 3: Geometry and TrigonometryIB Maths AA SL – Topic 4: Statistics and ProbabilityIB Maths AA SL – Topic 5: CalculusIB Maths AA HL – Topic 1: Number and AlgebraIB Maths AA HL – Topic 2: FunctionsIB Maths AA HL – Topic 3: Geometry and TrigonometryIB Maths AA HL – Topic 4: Statistics and Probability

All maths worksheets  ·  Full resource library  ·  Maths tuition with Mr. Ahmad

Get new maths worksheets by email

One short email when new worksheets are added. No spam, unsubscribe any time.



    Book a trial lesson (50% off)

    Try a full first lesson with Mr. Ahmad at 50% off before you commit to anything. It is a real lesson, not a sales call: bring a topic the student is stuck on and we will work through it together.

    • Online from anywhere in the world, or in person across Dubai
    • You get honest feedback on where the student stands and what to fix first
    • No contract — if it is not the right fit, that is fine

    Prefer to talk? Message on WhatsApp and you will hear back within 24 hours.



      Stuck on this topic?

      Book a trial lesson with Mr. Ahmad at 50% off and bring this worksheet with you — online worldwide or in person in Dubai.

      WhatsApp nowBook a trial lesson (50% off)
      Scroll to Top